7 TRACES, BITRACES, AND HUBERT ALGEBRAS 357 that there exists no trace on «£?(H) if H is an /«/zmte-dimensional Hilbert space (Problem at the end of this section). The corresponding notion for positive Hilbert forms is that of a bitrace. A bitrace on A is by definition a positive Hilbert form g on A which satisfies the additional condition (15.7.3) S(y*,x) = g(x9y). This reduces to (1 5.7.1 ) when g(x, y) =f(y*x), /being a positive linear form. Example (15J.4) Consider the algebra ^2(E) of Hilbert-Schmidt operators on an infinite-dimensional separable Hilbert space (15.4.8). For any Hilbert basis (#„) of E and any pair (u, v) of Hilbert-Schmidt operators on E, the sum £ (u(0n) | v(anj) is defined, because (u(an) \ v(an)) = ((v*u)(an) \ an\ and we have seen that the double family ((v*u)(an) \ am) is absolutely summable (15.4.8), since v*u is a Hilbert-Schmidt operator. Moreover, for each Hilbert basis (bn) of E, we have ) = I («(«„) I bm)(bn | t,(an)) and therefore (b J | This shows that the sum J] (u(an) \ v(an)) does not depend on the orthonormal n basis (an) originally chosen. If we denote this sum by g(u, v), then it is immediately seen that g is a positive hermitian form on J5?2(E), such that g(u,u)= \\u\\2- Also, if w, v, w are three Hilbert-Schmidt operators, then (u(v(an)) | w(aw)) = (r(aw) | u*(w(ajj)9 hence ^(wu, w) = g(v, M*W); and the cal- culation above shows that g(v*, w*) = #(w, y). Hence g is a bitrace on ^f 2(E). It can be shown that it does not come from a trace on this algebra (Problem). Note also that g satisfies conditions (U) and (N) of (15.6). For (U), this fol- lows from the inequality (15.4.8.2); for (N), observe that the endomorphisms of finite rank are dense in JS?2(E) (15.4.8) and that if w(E) = F is finite- dimensional then we may write u = PF o u (in the notation of (6.3.1)) where PF is also of finite rank.other hand, it is easily seen a (not necessarily